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WAEC 2010 · Paper 2 · Q4

The diagram shows a cone with slant height 10.5 cm10.5\text{ cm}. If the curved surface area of the cone is 115.5 cm2115.5\text{ cm}^2, calculate, correct to 3 significant figures, the: [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

10.5 cmhr
  1. (a)

    base radius, rr;

  2. (b)

    height, hh;

  3. (c)

    volume of the cone.

Worked solution (try it first)

(a)

  1. The curved surface area of a cone is πrl\pi r l.
  2. Put in the values: 227×r×10.5=115.5\frac{22}{7} \times r \times 10.5 = 115.5.
  3. 227×10.5=33\frac{22}{7} \times 10.5 = 33, so 33r=115.533r = 115.5.
  4. Divide: r=3.50 cmr = 3.50\text{ cm}.

(b)

  1. The radius, height and slant height form a right-angled triangle: h2=10.52−3.52h^2 = 10.5^2 - 3.5^2
    =110.25−12.25= 110.25 - 12.25
    =98= 98.
  2. So h=98h = \sqrt{98}
    =9.899= 9.899
    ≈9.90 cm\approx 9.90\text{ cm}.

(c)

  1. The volume of a cone is 13πr2h=13×227×3.52×9.899\frac13 \pi r^2 h = \frac13 \times \frac{22}{7} \times 3.5^2 \times 9.899.
  2. That is 13×38.5×9.899=127.0\frac13 \times 38.5 \times 9.899 = 127.0, so the volume is 127 cm3127\text{ cm}^3 to 3 significant figures.

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